Question: Solve for $z$, $ \dfrac{4z + 1}{z} = -\dfrac{6}{3z} + \dfrac{6}{z} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $z$ $3z$ and $z$ The common denominator is $3z$ To get $3z$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{4z + 1}{z} \times \dfrac{3}{3} = \dfrac{12z + 3}{3z} $ The denominator of the second term is already $3z$ , so we don't need to change it. To get $3z$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{6}{z} \times \dfrac{3}{3} = \dfrac{18}{3z} $ This give us: $ \dfrac{12z + 3}{3z} = -\dfrac{6}{3z} + \dfrac{18}{3z} $ If we multiply both sides of the equation by $3z$ , we get: $ 12z + 3 = -6 + 18$ $ 12z + 3 = 12$ $ 12z = 9 $ $ z = \dfrac{3}{4}$